Question

1③ Let Xn be {xm. a sequence and define nya) cos (ar). the sequence on = It is also known there are exactly strength 4 non-Zero values in and where they are the xn located sequence. in the Xn Tell me their sequence.

Answer 1

There are only 4 values in x[m] and they are non zero.If you see there is 2Ω term in Fourier transform Y(Ω) it means at both 2 and -2 x[m] has values.Since y(n) is the sum of x(m) values till n, x(m) must not have values before -2 because no other term of Ω is present. x(m)=0 for < -2. At n=-2 y(-2)=x(-2). And y(-2) is the coefficient of (2cos(2Ω)=e^-jΩ + e^jΩ.) So y(-2) = 0.5.

Also at -1 x(m) should have a value to make the sum zero.because -1Ω is not present in Y(Ω)

y(-1)= x(-2)+x(-1) =0. It implies x(-1)= -x(-2)= -y(-2)= -0.

x(0) and x(1) should be zero for y(0) =x(-2)+x(-1)+x(0)=0 and y(1)=0 because constant term and Ω term is absent in Y

And x(2)=0.5 for y(2) = 0.5=y(-2).

At 3 x(m) should have value to make the sum y(3) zero because no more terms of Ω are present.and this value of x(3)= -x(2)= -0.5.

And x(m) must be zero for m>3. To make sum y(n) zero.

So x(m) have values at m = -2, -1 ,2, 3 at which the values of x(m) are 0.5, -0.5, 0.5, -0.5 respectively.
yn) d(m) jan У(m) е m)eian Cos 22 2And

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